Fluid Mechanics and Hydraulic Machinery Miscellaneous
 Air flows at the rate of 1.5 m^{3}/s through a horizontal pipe with a gradually reducing crosssection as shown in the figure. The two crosssections of the pipe have diameters of 400 mm and 200 mm. Take the air density as 1.2 kg/m^{3} and assume inviscid incompressible flow. The change in pressure (p_{2 – p1) (in kPa) between sections 1 and 2 is}

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P_{1} + V_{1}² + Z_{1} = P_{2} + V_{2}² + Z_{2} [Z = constant] δg 2g δg 2g A_{1} = π × 0.4² = 0.1256 m² 2 A_{2} = π × 0.2² = 0.314 m² 4 P_{2} − P_{1} = V_{1}² − V_{2}² δg 2g P_{2} − P_{1} = 1 Q^{2} − Q^{2} 1.2 2 A_{1}² A_{2}² P_{2} − P_{1} = 0.6 × (1.5)² 1 − 1 A_{1}² A_{2}² P_{2} − P_{1} = 1.35 1 − 1 0.01577 0.00985
= 1.35 [63.41 – 1015.22]
= –1.28 ×10³ Pascal
= – 1.28 kPa
Correct Option: A
P_{1} + V_{1}² + Z_{1} = P_{2} + V_{2}² + Z_{2} [Z = constant] δg 2g δg 2g A_{1} = π × 0.4² = 0.1256 m² 2 A_{2} = π × 0.2² = 0.314 m² 4 P_{2} − P_{1} = V_{1}² − V_{2}² δg 2g P_{2} − P_{1} = 1 Q^{2} − Q^{2} 1.2 2 A_{1}² A_{2}² P_{2} − P_{1} = 0.6 × (1.5)² 1 − 1 A_{1}² A_{2}² P_{2} − P_{1} = 1.35 1 − 1 0.01577 0.00985
= 1.35 [63.41 – 1015.22]
= –1.28 ×10³ Pascal
= – 1.28 kPa
 Within a boundary layer for a steady incompressible flow, the Bernoulli equation

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Bernoulli equation does not hold because the flow is frictional.
Correct Option: D
Bernoulli equation does not hold because the flow is frictional.
 For a fluid element in a two dimensional flow field (xy plane), if it will undergo

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NA
Correct Option: C
NA
 Streamlines, path lines and streak lines are virtually identical for

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NA
Correct Option: C
NA
 A velocity field is given as
VV^{→} 3x²yî − 6xyzk̂
where x, y, z are in m and V in m/s. Determine if
(i) It represents an incompressible flow
(ii) The flow is irrotational
(iii) The flow is steady

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V = 3x²î − 6xyzk̂
∂u + ∂v + ∂w = 6xy + 0 − 6xy = 0 ∂x ∂y ∂z
Flow is incompressiblew_{y} = 1 ∂u − ∂w 2 ∂z ∂x = 1 [0 − (−6yz)] 2
w_{y} = 3yz
Flow is rotational. Flow field is independent of time since terms does not include time ‘t’.Correct Option: A
V = 3x²î − 6xyzk̂
∂u + ∂v + ∂w = 6xy + 0 − 6xy = 0 ∂x ∂y ∂z
Flow is incompressiblew_{y} = 1 ∂u − ∂w 2 ∂z ∂x = 1 [0 − (−6yz)] 2
w_{y} = 3yz
Flow is rotational. Flow field is independent of time since terms does not include time ‘t’.